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3 years ago

The 6 in the number 6,140 has a value __-- 648. times greater than the 6 in the number

Mathematics

1 answer:

nevsk [136]3 years ago

6 0

Answer:

1,000 because it's in the thousands place

Step-by-step explanation:

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Is this an irrational number?

GrogVix [38]

Answer:

no it is not an irrational number

8 0

2 years ago

Rotate RST 90° counterclockwise around the origin.<br>

Nostrana [21]

The image of the triangle RST when rotated 90° counterclockwise around the origin is (-15,5), (-15,15) and (-5,10)

<h3>
How to rotate the triangle?</h3>

The coordinates of RST are given as:

R = (5,15)

S = (15,15)

T = (10,5)

The rule of 90° counterclockwise around the origin is:

(x,y) -> (-y,x)

So, we have:

R' = (-15,5)

S' = (-15,15)

T' = (-5,10)

Hence, the image of the triangle when rotated 90° counterclockwise around the origin is (-15,5), (-15,15) and (-5,10)

Read more about rotation at:

brainly.com/question/4289712

#SPJ1

4 0

2 years ago

A coin is to be tossed as many times as necessary to turn up one head. Thus the elements c of the sample space C are H, TH, TTH,

slavikrds [6]

Answer:

Step-by-step explanation:

As stated in the question, the probability to toss a coin and turn up heads in the first try is $\frac{1}{2}$ , in the second is $\frac{1}{4}$ , in the third is $\frac{1}{8}$ and so on. Then, P(C) is given by the next sum:

$P(C)=\sum^{\infty}_{n=1}(\frac{1}{2} )^{n}=1$

This is a geometric series with factor $\frac{1}{2}$ . Then is convergent to $\frac{1}{1-\frac{1}{2}}-1=1.$ . With this we have proved that P(C)=1.

Now, observe that

$P(H)=\frac{1}{2}, P(TH)=\frac{1}{4},P(TTH)=\frac{1}{8},P(TTTH)=\frac{1}{16},P(TTTTH)=\frac{1}{32},P(TTTTTH)=\frac{1}{64}.$

Then

$P(C1)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} =\frac{31}{32}$

$P(C2)=P(TTTTH)+P(TTTTTH)=\frac{1}{32}+\frac{1}{64} =\frac{3}{64}$

$P(C1\cap C2)=P(TTTTH)=\frac{1}{32}$

and

$P(C1\cup C2)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)+P(TTTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} +\frac{1}{64}=\frac{63}{64}$